Laws Of Limits

We can calculate the limits of a function from its graph as shown in the following
example.

Illustrative Example

All the following statements about the function in the graph above are true.

At x = -2 :

limf (x) = 5
x -> -2+

At x = 0 :

limf (x) = 1
x -> 0-

limf (x) = 1
x -> 0+

limf (x) = 1
x -> 0

Also f (1) = 1

At x = 1 :

limf (x) = 1
x -> 1-

Even though f (1) = 2

limf (x) = 2
x -> 1+

Hence limf (x)
does not exist as the One-sided limits are not equal.
x -> 1+

At x = 3 :

limf (x) = 6
x -> 3-

= lim f (x) x -> 1+

Hence lim f (x)
= 6 even though f (x) = 3
x -> 3

At x = 5 :

Hence lim f (x)
= -2
x -> 5-

At every other point 'a' between -2 and 5, f (x) has a limit as x -> a.

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Limit Laws

Suppose 'c' is a constant and also lim f (x) and
lim g (x) both exist then the following rules
hold true in the calculation of limits.
x -> a
x -> a

From the Product rule we can also conclude by repeated multiplication that:

(Power Rule)

lim [f
( x )]n x -> a

=

[ lim f
( x )]n x ->
a

where n is positive integer.

Similarly we have

(Root Rule)

lim n
√f ( x ) x -> a

=

n
√ lim
f ( x ) x -> a

where n is positive integer and if it is even then we assume that lim
f ( x ) > 0.x -> a

Illustrative Examples:

lim x 2 x -> 2

=

( lim x )x -> 2

( lim x )x -> 2

.............

Product Rule

=

( 2 )

( 2 )

.............

Basic Substitution Rule

=

4

Therefore,

lim x 2 x -> 2

=

4

lim 3x2 = 3 lim x2 = 3
( 4 ) = 12
x -> 2x -> 2

lim
x -> 2

x 2 - 2

3x2

=

lim ( x2 - 2 )
x -> 2
lim 3x2x -> 2

.................. (Quotient Rule)

=

lim x2
- lim 2
x -> 2x -> 212

............... (Subtraction Rule)

=

4 - 2
12

............... (Basic Rule)

=

1
6

Conclusion: (Direct Substitution Property)

If f (x) is any polynomial function then

lim f (x) = f (a)x -> 2

Also if g (x) is also a polynomial and g (a)
≠ 0 then

lim
x -> a

f ( x )

g ( x )

=

f ( a )

g ( a )

Examples:

lim
x -> 1

x3 + 4x2 + 3x + 2

6x - 1

=

13 + 4
(1)2 + 3 (1) + 2

6 (1) - 1

=

10

5

=

2

lim
x -> 2

x2 - 4

x - 2

Here we cannot apply the substitution as it takes 0 / 0 form nor can we apply quotient
rule as the denominator x - 2 = 0 for x = 2.
Instead let us apply some algebra before taking the limit.

x2
- 4

x - 2

=

(x + 2) (x - 2)

(x - 2)

Now we know that x -> 2 but x ≠ 2. Hence x - 2 ≠ 0 and
we can cancel it off.

lim
x -> 2

x2
- 4

x - 2

=

lim
x -> 2

(x + 2) (x - 2)

(x - 2)

=

lim (x + 2)
x -> 2

= 2 + 2 = 4

Therefore,

lim
x -> 2

x2
- 4

x - 2

=

4

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